Comps Study Guide for Linear Algebra

Comps Study Guide for Linear Algebra Department of Mathematics and Statistics September, 2017 This study guide was written to help you prepare for the linear algebra portion of the Comprehensive and Honors Qualifying Examination in Mathematics. It is based on the Syllabus for the Comprehensive Examination in Linear Algebra (Math 271/272) available on the Department website. Each topic from the syllabus is accompanied by a brief discussion and examples from old exams. When reading this guide, you should focus on three things: • Understand the ideas. If you study problems and solutions without understanding the underlying ideas, you will not be prepared for the exam. • Understand the strategy of each proof. Most proofs in this guide are short—the hardest part is often knowing how to start. Focus on the setup step rather than falling into the trap of memorizing proofs. • Understand the value of scratchwork. Sometimes scratchwork is needed to explore possible approaches to a computation or proof. That is, sometimes it is helpful to work out some details on the side by trial and error before writing up a clear, presentable solution. The final section of the guide has some further suggestions for how to prepare for the exam. 1 Vector Spaces and Subspaces Some basic things to be aware of, although they do not arise directly on the comps: • The definition of a vector space: A set with operations called addition and scalar multiplication (by elements of R) satisfying a certain long list of axioms. • One of the equivalent definitions of a subspace: as a subset W of a vector space V such that W is a vector space in its own right, with the same addition and scalar multiplication operations. The elements of a vector space V are called vectors, even if those elements are functions, matrices, or other objects. The additive identity of V is called the zero vector, and it is usually denoted 0 or 0V . Simple vector space examples. Here are some vector spaces you should know, each with standard addition and scalar multiplication operations. You don’t need to memorize the notation (other than for Rn , which you surely already know), because any such notation will be defined on the exam if it appears. • Rn is the vector space of ordered n-tuples of real numbers. Sometimes denoted Rn . Sometimes its elements are written as row vectors (x1 , . . . , xn ) and sometimes as columns. Note: dim(Rn ) = n. • Pn (R) is the vector space of polynomials of degree less than or equal to n. Sometimes denoted Pn or P n or Pn or some other such variation. Note: dim(Pn (R)) = n + 1. • Mm×n (R) is the vector space of m × n matrices with real entries. Sometimes denoted Mm×n or some other such variation. Note: dim(Mm×n (R)) = mn. 1 There are also some infinite-dimensional vector spaces which arise occasionally. For example, P (R) is the vector space of all polynomials, F (R) is the vector space of all functions f : R → R, and C(R) is the vector space of all continuous functions f : R → R. Again, don’t bother memorizing the notation, which will be fully defined if it appears; just be ready to work with such spaces if they show up on the exam. Subspace Theorem. The following Theorem is usually used to check whether a given subset is a subspace; in fact, some books use it as the definition of a subspace. Theorem. Let V be a vector space. A subset W ⊆ V is a subspace if it satisfies the following properties: 1. W 6= ∅. 2. For all x, y ∈ W and all c ∈ R, we have cx + y ∈ W . Property 1 is usually verified by proving that 0V ∈ W . Property 2 can be replaced by two separate statements: Closed under addition: For all x, y ∈ W , we have x + y ∈ W . Closed under scalar multiplication: For all x ∈ W and c ∈ R, we have cx ∈ W . That is, a subset W of a vector space V is a subspace if and only if W is nonempty, closed under addition, and closed under scalar multiplication. Note: • The empty set ∅ is not a vector space. Instead, the smallest vector space is the trivial space, {0}. • Every vector space V has two obvious subspaces: the trivial subspace {0} ⊆ V , and the improper subspace V ⊆ V . (Obviously, the two coincide if and only if V = {0} is trivial.) 1 (March 2006) Let U and V be subspaces of a vector space W . (a) Prove that U ∩ V is a subspace of W . (b) Prove that U + V = {u + v : u ∈ U, v ∈ V } is a subspace of W . (c) Give an example to show that U ∪ V need not be a subspace of W . Proof. (a): (Nonempty) We have 0W ∈ U and 0W ∈ V , since both U and V are subspaces. Hence 0W ∈ U ∩ V . (Closure) Given x, y ∈ U ∩V and c ∈ R, we have cx +y ∈ U and cx +y ∈ V since both are subspaces. Hence cx + y ∈ U ∩ V . QED (a) (b): (Nonempty) We have 0W ∈ U and 0W ∈ V , since both are subspaces. Hence 0W + 0W ∈ U + V . (Closure) Given x, y ∈ U + V and c ∈ R, there exist u1 , u2 ∈ U and v1 , v2 ∈ V such that x = u1 + v1 and y = u2 + v2 . Thus, cx + y = c(u1 + v1 ) + (u2 + v2 ) = (cu1 + u2 ) + (cv1 + v2 ) ∈ U + V QED (b) (c): Let W = R2 , let U = Span({(1, 0)}), and let V = Span({(0, 1)}). [That is, U is the x-axis, and V is the y-axis.] Then U and V are subspaces because they are each the span of a set. However, (1, 0) ∈ U ⊆ U ∪ V and (0, 1) ∈ V ⊆ U ∪ V , but (1, 0) + (0, 1) = (1, 1) 6∈ U ∪ V . Thus, U ∪ V is not closed under addition and hence is not a subspace. QED (c) Comment 1. In the “Nonempty” step of (b), there was no need to observe that 0W + 0W = 0W . Yes, that’s true, and it wouldn’t hurt to say it, but it would be unnecessary. We’re simply trying to show that U + V 6= ∅, so all we need to do is produce an element; we aren’t required to simplify that element. On the other hand, in part (c), it was important to simplify the sum (1, 0) + (0, 1), to verify that the result is not an element of U ∪ V . Comment 2. In part (c), we made use of Span (to be discussed below) because it made the proof shorter, rather than verifying by hand that both U and V are subspaces. Also in part (c), there are many ways to do this. (Any choice of W with any subspaces U and V will work, as long as U 6⊆ V and V 6⊆ U .) But it’s generally best to pick as simple an example as you can. 2 Linear combinations, etc. Let V be a vector space, and let S = {v1 . . . , vn } ⊆ V be a finite set of vectors in V . • A linear combination of elements of S is an expression a1 v1 +· · ·+an vn for some scalars a1 , . . . , an ∈ R. • The span of S, denoted Span(S), is the set of all linear combinations of elements of S. That is, – – – – Span(S) = {a1 v1 + · · · + an vn | a1 , . . . , an ∈ R}. Fact: Span(∅) = {0}. Fact: Span(S) is a subspace of V . Fact: If W is a subspace and S ⊆ W , then Span(S) ⊆ W . If Span(S) = W , we say that S spans W . • The set S is linearly dependent if there exist scalars a1 , . . . , an ∈ R that are not all zero such that a1 v1 + · · · + an vn = 0. S is linearly independent if it is not linearly dependent. Equivalently, for any scalars a1 , . . . , an ∈ R such that a1 v1 + · · · + an vn = 0, we must have a1 = · · · = an = 0. • The set S ⊆ V is a basis for V if S is linearly independent and Span(S) = V . • The dimension of V is the number dim(V ) of elements in a basis for V . (It is a Theorem that any two bases for V have the same number of elements.) If V has no finite basis, we say dim(V ) = ∞. Side notes: Linear independence really should include an extra specification that we first ensure v1 , . . . , vn are all distinct; but this subtlety does not arise on the comps. Also, there are notions of linear combinations, span, and linear (in)dependence for an infinite set S ⊆ V , but again, these do not arise on the comps. 2 (March 2007) Suppose that V is a vector space and {v1 , . . . , vn } ⊆ V is linearly independent. Also assume that v ∈ V is not contained in the span of {v1 , . . . , vn }. Prove that {v, v1 , . . . , vn } is linearly independent. Proof. Given scalars a0 , a1 , . . . , an ∈ R such that a0 v + a1 v1 + · · · + an vn = 0, we need to show that a0 = a1 = · · · = an = 0. Case 1 : Suppose a0 = 0. Then a1 v1 + · · · + an vn = 0, and hence a1 = · · · = an = 0, because {v1 , . . . , vn } is linearly independent. Thus, a0 = a1 = · · · = an = 0, as desired. Case 2 : Suppose a0 6= 0. Then a0 v = −a1 v1 − · · · − an vn , and multiplying both sides by 1/a0 (which a1 an is a real number, since a0 6= 0), we have v = − v1 −· · ·− vn ∈ Span({v1 , . . . , vn }), contradicting a0 a0 the hypotheses. Thus, Case 2 cannot happen, and we are done by Case 1. QED Comment 1. Don’t start a proof by regurgitating what the hypotheses mean. Instead, focus on the statement you’re asked to deduce: for this problem, that the set {v, v1 , . . . , vn } is linearly independent. So our first line should be to suppose we are given scalars ai for which a0 v+a1 v1 +· · ·+an vn = 0, and our last line should be to conclude that a0 = a1 = · · · = an = 0. Once you know the first and last line, then you can go back and try to figure out how the hypotheses could help you get there. Comment 2. In this problem, two cases are required. We discovered that by doing some scratchwork, and realizing that the span hypothesis could only be used in one situation, and the linear independence hypothesis can be used in the other. Scratchwork is important! Row reduction. Many problems in linear algebra end up requiring row reduction. You will need to know: • How to set up a system of equations when necessary. • The three kinds of row reduction steps. (Switching two rows, multiplying a row by a nonzero scalar, and adding a multiple of one row to another.) • The overall strategy, leading to a matrix in echelon form. • How to interpret the echelon form to solve your problem. 3 At the most basic level, you solve a linear  x1 + 2x2 + 2x4 =2 2x2 + 4x2 + x3 + 7x4 = 5 =⇒  x1 + 2x2 − 2x3 − 4x4 + x5 = 3 system by row reducing its augmented   1 1 2 0 2 0 2 2 4 1 7 0 5  → ··· →  0 1 2 −2 −4 1 3 0 matrix:  2 0 2 0 2 0 1 3 0 1 . 0 0 0 1 3 You should do this row reduction yourself (see 6 for a worked out example). The echelon matrix above has pivots [first nonzero entries of rows of an echelon form] in columns 1, 3 and 5. Thinking of the first five columns as corresponding to the variables, we see that x2 and x4 are free variables [since there are no pivots in these columns]. So the system of equations given by the echelon form is straightforward to solve: x1 + 2x2 + 2x4 x3 + 3x4 =2 =1 x5 = 3 x1 = 2 − 2x2 − 2x4 x3 = 1 − 3x4 x5 = 3 =⇒ Notice how we solve for the pivot variables x1 , x3 , x5 in terms of the free variables. So the general solution is           −2 −2 2 2 − 2x2 − 2x4 x1  0  1  0 x2   x 2           x3  =  1 − 3x4  = 1 + x2  0 + x4 −3 , where x2 , x4 ∈ R are arbitrary.            1  0  0 x4   x4 0 0 3 3 x5 On the linear algebra comps, one often encounters homogeneous systems, where all the constant terms are zero. In this example, the corresponding homogeneous system is x1 + 2x2 + 2x4 =0 2x2 + 4x2 + x3 + 7x4 =0 x1 + 2x2 − 2x3 − 4x4 + x5 = 0 with row reduction and general solution  1 2  2 4 1 2   0 2 0 0 1 1 7 0 0  −→  0 −2 −4 1 0 0 2 0 0 0 1 0 2 3 0 0 0 1       −2 x1 −2  0 x2   1 0            0  =⇒  x3  = x2  0 + x4 −3 .  1     0 0 x4 0 0 x5  We will say more about matrices in Section 3 of this study guide. For now, we mention two important uses of row reduction and the echelon form: • The nullspace of a matrix A is the solution set of its corresponding homogeneous system of equations. The method above produces a basis of the nullspace of A, namely the set of vectors that the free variables end up multiplied by in the solution. That is, in the example above,     −2   −2            1 2 0 2 0 1    0   0 , −3 . 1 7 0 has basis the nullspace of 2 4       1 2 −2 −4 1  0  1      0 0 • The column space of a matrix A is the span of its columns. If B is the echelon form of A, then the columns of A corresponding to the columns of B with pivots form a basis of the column space. In the example above, the echelon form has pivots in columns 1, 3 and 5, so that         1 2 0 2 0 0 0   1 2 ,  1 , 0 . 1 7 0 has basis the column space of 2 4   1 2 −2 −4 1 1 −2 1 See 6 for a similar problem, and see Section 2 for more on nullspaces and column spaces. 4 Row reduction arises most obviously when you are asked to solve an specific system of equations. But it can also arise at other times, especially on problems that involve explicit vectors. Here are some examples of problems involving linear independence and span where row reduction arises naturally. 3 (March 2016) Is the following set of polynomials linearly independent? Explain your answer: {x3 − 3x + 1, x2 + 2x + 2, x3 − 2x2 }. Solution/Proof. Yes. Call the three polynomials f , g, h. Given scalars a, b, c ∈ R such that af + bg + ch = 0, we have (a + c)x3 + (b − 2c)x2 + (−3a + 2b)x + (a + 2b) = 0, i.e., a + c = 0, b − 2c = 0, −3a + 2b = 0, a + 2b = 0.   1 0 1 0  0 1 −2 0  . Thus, (a, b, c) is a solution of the system of equations described by   −3 2 0 0  1 2 0 0   1 0 1 0  0 1 −2 0  . Row reduction [omitted here; try it yourself!] leads to the echelon form   0 0 1 0  0 0 0 0 There are no free variables. The third row of the echelon form gives c = 0, so the second gives b = 0, and the third gives a = 0. Since the only solution (a, b, c) to af + bg + ch = 0 is a = b = c = 0, the set {f, g, h} is linearly independent. QED 4 (January 2015) Suppose that V is a vector space and u and v are vectors in V . Show that Span({3u + v, u − v}) = Span({u, v}). Proof. (⊆): Clearly, 3u + v ∈ Span({u, v}), and u − v ∈ Span({u, v}). Thus, since Span({u, v}) is a subspace, we have Span({3u + v, u − v}) ⊆ Span({u, v}). (⊇): Note that 4u = (3u + v) + (u − v), so u = 14 (3u + v) + 14 (u − v) ∈ Span({3u + v, u − v}). Similarly, 4v = (3u + v) − 3(u − v), so v = 41 (3u + v) − 43 (u − v) ∈ Span({3u + v, u − v}). Since Span({3u + v, u − v}) is a subspace, we have Span({3u + v, u − v}) ⊇ Span({u, v}). QED Comment. The (⊇) direction required some seemingly clever choices of coefficients. How did we find them? By doing scratchwork! For example, when showing u ∈ Span({3u + v, u − v}), we needed   3 1 1 to find scalars a, b ∈ R so that a(3u + v) + b(u − v) = u. This becomes the system , 1 −1 0 which we solved (in scratchwork) to get a = 14 and b = 14 . 17 Be sure you are comfortable with row reduction and solving linear systems. See problems , 20 , 21 , 22 and 23 of this study guide for more problems that require row reduction. 6 , 7 , 16 , Side Note: A matrix B is in echelon form if it fits the staircase pattern, where each pivot has all zeros below it and to the left of it. B is in reduced echelon form if, in addition, each pivot also has all zeros above it, and each pivot is exactly 1. Some of the echelon forms in this guide are reduced (e.g., in the example on page 4), and some are not (e.g., in Example 3 ). On the comps, you may use either kind, as you prefer. On the one hand, the reduced echelon form can make it easier to write down the solution; on the other hand, if you only need the number or location of the pivots, a non-reduced echelon form suffices. 5 2 Linear Transformations Let V and W be vector spaces, and let T : V → W be a function. We say T is a linear transformation (or a linear map, or simply that T is linear ) if for all x, y ∈ V and all c ∈ R, we have T (cx + y) = cT (x) + T (y). • The linearity property can be replaced by two separate statements: Respects addition: For all x, y ∈ V , we have T (x + y) = T (x) + T (y). Respects scalar multiplication: For all x ∈ V and c ∈ R, we have T (cx) = cT (x). • If U : W → X is another linear transformation, then the composition U ◦ T : V → X is also linear. The composition U ◦ T is often denoted simply U T . • Don’t confuse proving “T is linear” with “W is a subspace.” They are similar but not the same. (In particular, proving W is a subspace also requires proving it’s nonempty.) • Sometimes we drop the parentheses and write T x instead of T (x). • When T is linear, the following identity can be very useful: T (a1 v1 + · · · + an vn ) = a1 T (v1 ) + · · · + an T (vn ). • If A ∈ Mm×n (R) is an m × n matrix, then the function T : Rn → Rm by T (x) = Ax is linear. • If T : Rn → Rm is linear, then there is a matrix A ∈ Mm×n (R) so that T is given by T (x) = Ax. Kernel, image, nullity, and rank. Let T : V → W be a linear transformation. The kernel or nullspace of T is {v ∈ V | T (v) = 0W } ⊆ V. It is usually denoted either Ker(T ) or N (T ), or sometimes nullspace(T ). Its dimension dim(Ker(T )) is called the nullity of T , sometimes denoted nullity(T ). The image or range of T is {T (v) | v ∈ V } = {w ∈ W | ∃v ∈ V s.t. T (v) = w} ⊆ W. It is usually denoted either Im(T ) or R(T ), or sometimes range(T ). Its dimension dim(Im(T )) is called the rank of T , sometimes denotes rank(T ). If T : Rn → Rm is multiplication by the matrix A, the range of T is sometimes called the column space of A, because it is precisely the span of the columns of A. • Fact: N (T ) is a subspace of V , and R(T ) is a subspace of W . • Fact: T is one-to-one if and only if N (T ) = {0V }, i.e., iff N (T ) is as small as possible, i.e., iff nullity(T ) = 0. (Recall f : X → Y is one-to-one means for all x1 , x2 ∈ X with f (x1 ) = f (x2 ), we have x1 = x2 .) • Fact: T is onto if and only if R(T ) = W , i.e., iff R(T ) is as big as possible. If dim(W ) < ∞, this is equivalent to saying rank(T ) = dim(W ). (Recall f : X → Y is onto means for all y ∈ Y , there exists x ∈ X with f (x) = y.) • Fact: If T : Rn → Rm is represented by a matrix A, then N (T ) = {x ∈ Rn | Ax = 0} and R(T ) = column space of A. Furthermore, after finding an echelon form of A via row reduction, rank(T ) is the number of columns with pivots (since the corresponding columns form a basis of R(T )), and nullity(T ) is the number of columns without pivots (since the vectors that give the general solution form a basis of N (T )). • Given a linear map T : V → W , know how to use row reduction to find bases for both N (T ) and R(T ). Also be able to decide whether or not T is one-to-one, and whether or not T is onto. 6 Some proof problems combine the notions of linear maps, span, linear independence, one-to-one, and/or onto, so be prepared for problems like the following. 5 (March 2013) Suppose that V and W are vector spaces, T : V → W is a linear transformation, and let v1 , v2 , . . . , vn ∈ V . Prove that if {T (v1 ), T (v2 ), . . . , T (vn )} spans W and T is one-to-one, then {v1 , v2 , . . . , vn } spans V . Proof. Given v ∈ V , we need to show v ∈ Span({v1 , . . . , vn }). Since T (v) ∈ W = Span({T (v1 ), . . . , T (vn )}, there exist scalars a1 , . . . , an ∈ R such that T (v) = a1 T (v1 ) + · · · + an T (vn ) = T (a1 v1 + · · · + an vn ). Since T is one-to-one, we have v = a1 v1 + · · · + an vn ∈ Span({v1 , . . . , vn }). QED Comment 1. As always in a proof, start by focusing on the goal. We need to show that {v1 , v2 , . . . , vn } spans V ; so ask yourself what the first and last line of such a proof should be. Comment 2. If you know T is linear, then there’s a good chance that you will have to use the fact that T (a1 v1 + · · · + an vn ) = a1 T (v1 ) + · · · + an T (vn ). So be ready to do so. Here is a problem about finding bases for the column space and null space of a particular matrix. (March  1 2 A = 1 2 2 4 6 2014)Find [bases for] the column space and the null space (or kernel) for the matrix 2 3 1 2. 3 5 Solution. We  1 −R1 1 −2R1 2 do row reduction:   2 2 3 +2R2 1 0 2 1 2 → 4 3 5 −R2 0 2 0 0 2 −1 −1   3 1 −1 → ×(−1) 0 −1 0 2 0 0 0 −1 0   1 1 −1 → 0 0 0 2 0 0 0 1 0  1 1 0 This echelon form matrix has pivots in columns 1 and 3. So those columns of the original matrix A are a basis for the column space. That is, (column space of A) = Span{(1, 1, 2), (2, 1, 3)}. For the kernel, the free variables are x2 , x4 , and the echelon form gives the general solution       x1 −2 −1 x2        = x2  1 + x4  0 =⇒ Ker(A) = Span{(−2, 1, 0, 0), (−1, 0, −1, 1)}. x3   0 −1 x4 0 1 Comment 1. When you do row reduction, be sure to indicate what operations you are doing at each step. Do this both to avoid errors and to avoid losing excess points if you do make an error. Comment 2. Don’t do two row reduction operations in one step unless those two operations commute. For example, you can add multiples of row 1 to both rows 2 and 3 in one step, but you can’t add row 1 to row 2 and multiply row 2 by −1 in the same step. Sometimes you may be asked to find a basis for some subspace when there is no obvious linear map or matrix. In that case, usually the subspace is actually the kernel or image of some linear map or matrix which will make itself apparent once you starting playing with it. Here is an example of such a problem. 7 7 (March 2009) Let V be the vector space of polynomials with real coefficients and of degree at most 3, and let W = {f ∈ V | f (0) = f 00 (0) and f 0 (1) = 0}. (a) Prove that W is a subspace of V . (b) Find a basis for W . (a): Proof. (Nonempty) Let f = 0 ∈ V . Then f 0 = 0 and f 00 = 0, so f (0) = 0 = f 00 (0) and f 0 (0) = 0. That is, f ∈ W . So W 6= ∅. (Closure) Given f, g ∈ W and c ∈ R, we have (cf + g)(0) = cf (0) + g(0) = cf 00 (0) + g 00 (0) = (cf + g)00 (0) and (cf + g)0 (1) = cf 0 (1) + g 0 (1) = c · 0 + 0 = 0, and hence cf + g ∈ W . QED (a) 2 3 0 (b): Solution. Write an element f of V = P3 (R) as f = a + bx + cx + dx . Then f = b + 2cx + 3dx2 , and f 00 = 2c + 6dx. So f (0) = a, f 00 (0) = 2c, and f 0 (1) = b + 2c + 3d. Hence, W = {a + bx + cx2 + dx3 ∈ V | a = 2c and b + 2c + 3d = 0}. Thus, we are solving  the system a − 2c = 0 and b + 2c + 3d = 0 for a, b, c, d ∈ R. That is, we need 1 0 −2 0 the kernel of A = , which is already in echelon form. Since there are pivots only in 0 1 2 3 the columns for a and b, the free variables are c and d. Setting c = 1 and d = 0 gives a = 2 and b = −2, yielding x2 − 2x + 2. Setting c = 0 and d = 1 gives a = 0 and b = −3, yielding x3 − 3x. So {x2 − 2x + 2, x3 − 3x} is a basis for W . Comment. In the moment, it can be easy to forget that we’re dealing with elements of V , which are polynomials. So when you finish the problem, make sure your answer actually fits. In particular, the answer to (b) is not {(2, −1, 1, 0), (0, −3, 0, 1)}, but rather {x2 − 2x + 2, x3 − 3x}. Some facts about dimension. • Fact: If X ⊆ V is a subspace, then dim(X) ≤ dim(V ). Moreover, if dim(V ) < ∞, then dim(X) = dim(V ) if and only if X = V . • Fact: Let V be a vector space with dim(V ) = n, and let S ⊆ V be a set of m distinct vectors in V . – If m < n, then S cannot span V . – If m > n, then S cannot be linearly independent. Here are two useful facts that might be called “two-out-of-three” theorems: • Theorem: Let V be a vector space, and let S ⊆ V be a set of n distinct vectors in V . If any two of the following conditions hold, then all three hold (and S is a basis for V ): 1. S is linearly independent. 2. S spans V . 3. dim(V ) = n. • Theorem: Let T : V → W be a linear transformation, and suppose that at least one of V, W is finite-dimensional. If any two of the following conditions hold, then all three hold: 1. T is one-to-one. 2. T is onto. 3. dim(V ) = dim(W ). In this case, T is invertible. (More on this later; see page 13 of this Guide.) Here is a very important fact, sometimes called the Rank-Nullity Theorem or the Dimension Theorem. • Theorem: Let T : V → W be a linear transformation. Then rank(T ) + nullity(T ) = dim(V ). 8 8 (January 2012) Let C be a 3 × 5 real-valued matrix. Answer the following questions about C and briefly justify your answers: (a) Can the columns of C be linearly independent? (b) Does the equation Cx = 0 have a unique solution with x ∈ R5 ? (c) Assume that the span of the columns of C is all of R3 . Can you determine the nullity (=dimension of the null space or kernel) of C? Answers. (a): No, the columns of C must be linearly dependent. The columns of C are five vectors in R3 . Since dim(R3 ) = 3 < 5, any set of five vectors in R3 is linearly dependent. (b): No, the equation Cx = 0 has infinitely many solutions. Since the range R(C) is a subspace of R3 , we have rank(C) ≤ 3. Since multiplication-by-C is a linear map from R5 to R3 , the Rank-Nullity Theorem yields nullity(C) = dim(R5 ) − rank(C) ≥ 5 − 3 = 2. Thus, the solution set N (C) to Cx = 0 is a subspace of R5 of dimension at least 2, so it has infinitely many elements. (c): Yes, we must have nullity(C) = 2. The rank of C is by definition rank(C) = dim(R(C)). Since the range R(C) is the span of the columns of C, the hypothesis says that R(C) = R3 , and hence rank(C) = 3. Thus, the Rank-Nullity Theorem yields nullity(C) = dim(R5 ) − rank(C) = 5 − 3 = 2. Comment. An alternate way to do this problem is to say that C has some echelon form B, with some number of pivots between 0 and 3. (Max of 3, since there are 3 rows.) Then consider what the columns and rows with or without pivots say about each of parts (a), (b), (c). 9 (January 2009) Let A be an n × n matrix with real entries such that A2 = 0. (a) Prove that the column space of A is contained in the nullspace of A. (Recall that the nullspace, or kernel, is the set of vectors v ∈ Rn satisfying Av = 0.) (b) Prove that rank(A) ≤ nullity(A). (Recall that the rank is the dimension of the column space and the nullity is the dimension of the nullspace.) (c) Prove that nullity(A) ≥ 12 n. Proof (a): Write R(A) for the column space of A and N (A) for the nullspace. We must show R(A) ⊆ N (A). Given w ∈ R(A), there exists v ∈ Rn such that w = Av. Thus, Aw = A(Av) = (A2 )v = 0v = 0, and hence w ∈ N (A). QED (a) (b): Since R(A) ⊆ N (A) by part (a), we have rank(A) = dim(R(A)) ≤ dim(N (A)) = nullity(A). QED (b) (c): The Rank-Nullity Theorem for A gives n = dim(Rn ) = rank(A) + nullity(A) ≤ 2 nullity(A), where the inequality is from part (b). Dividing both sides by 2, we have nullity(A) ≥ 9 1 n. QED (c) 2 (March 2011) Suppose that T : R3 → R2 and U : R2 → R3 are linear transformations, and for all v ∈ R2 , (T ◦ U )(v) = v. 10 (a) Prove that T is onto. (b) What is the nullity of T ? Is T one-to-one? Justify your answers. Proof. (a): Given v ∈ R2 , let w = U (v) ∈ R3 . Then T (w) = T (U (v)) = (T ◦ U )(v) = v. QED (a) (b): We have nullity(T ) = 1. No, T is not one-to-one. To see this, note by part (a) that the range R(T ) is all of R2 , because T is onto. Thus, the rank of T is rank(T ) = dim(R(T )) = 2. So by the Rank-Nullity Theorem, nullity(T ) = dim(R3 ) − rank(T ) = 3 − 2 = 1. Since the nullspace N (T ) has dimension nullity(T ) = 1 > 0, T is not one-to-one. QED (b) 11 (February 2007) Let V and W be vector spaces over R and let T : V → W be a linear map. Assume that V has a basis {v1 , . . . , vn } such that {T (v1 ), . . . , T (vn )} spans W . (a) Prove that T is onto. (b) If dim V = dim W , what else can you conclude about T ? Explain your reasoning. Proof. (a): Given w ∈ W , there exist a1 , . . . , an ∈ R such that w = a1 T (v1 ) + · · · + an T (vn ), since {T (v1 ), . . . , T (vn )} spans W . Thus,
w = T a1 v1 + · · · + an vn lies in the image of T , as desired. QED (a) (b): T is one-to-one and onto (and hence invertible). We know from part (a) that T is onto, and we know from the hypothesis of (b) that dim(W ) = dim(V ) = n < ∞. Thus, by a two-out-of-three theorem, T must also be one-to-one. QED (b) 12 (January 2012) Let T : V → V be a linear transformation on a finite dimensional vector space V . Suppose T is one-to-one (injective). Prove that if {v1 , . . . , vn } is a basis for V , then {T (v1 ), . . . , T (vn )} is also a basis for V . Proof. First, we claim that {T (v1 ), . . . , T (vn )} is linearly independent. Given scalars a1 , . . . , an ∈ R such that a1 T (v1 ) + · · · + an T (vn ) = 0, we have
T a1 v1 + · · · + an vn = 0. Since T is one-to-one, it has trivial kernel, so a1 v1 + · · · + an vn = 0. Because {v1 , . . . , vn } is linearly independent, it follows that a1 = · · · = an = 0, proving our claim. Now dim(V ) = n (because by hypothesis, it has a basis consisting of n elements), and by our claim, {T (v1 ), . . . , T (vn )} is a linearly independent set of n elements in V . Thus, by a two-out-of-three theorem, {T (v1 ), . . . , T (vn )} also spans V and hence is a basis for V . QED 10 3 Matrices Know: • an m × n matrix is an m-row, n-column grid of real numbers. • how to add two m × n matrices. • how to multiply AB, where A ∈ Mm×n (R) and B ∈ Mn×p (R). • matrix addition is commutative and associative. • matrix multiplication is associative but not commutative. • the m × n matrix of zeros (denoted 0 or O) is the additive identity: A + O = O + A = A. • the n × n identity matrix I, sometimes denoted In , has 1’s down the diagonal and 0’s everywhere else. It satisfies AI = A and IB = B (for A is m × n and B is n × p). • the distributive laws A(B + C) = AB + AC and (A + B)C = AC + BC. • How to convert a system of linear equations to a matrix equation of the form Ax = b. In addition, if A is a square matrix (i.e., n × n), know • A is a diagonal matrix if every entry not on the diagonal is 0. • The transpose of A, denoted At or AT , is the n × n matrix formed by reflecting A across the diagonal. • The determinant of A, denoted det(A), is a real number given by a more complicated formula. Know: – How to compute det(A) for a given n × n matrix, for any n ≥ 1. – The fact that det(AB) = det(A) det(B). – The fact that det(A) 6= 0 if and only if A is invertible (see later). – An alternate notation for a determinant is to put vertical lines instead of brackets around the  a b a b means the same thing as det , namely ad − bc. entries of the matrix. For example, c d c d Determinants don’t usually arise directly on comps exams, but you will need determinants to find eigenvalues. Matrix of a linear transformation. Let V and W be finite-dimensional vector spaces with bases α = {v1 , . . . , vn } and β = {w1 , . . . , wm }, respectively, and let T : V → W be a linear transformation. Find real numbers aij , for 1 ≤ i ≤ m and 1 ≤ j ≤ n, by computing T (v1 ), . . . , T (vn ) and expressing each as a linear combination of w1 , . . . , wm . That is, T (v1 ) = a11 w1 + a21 w2 + · · · + am1 wm T (v2 ) = a12 w1 + a22 w2 + · · · + am2 wm .. . T (vn ) = a1n w1 + a2n w2 + · · · + amn wm . Then, turning the grid of coefficients sideways, the  a11 a12  a21 a22  [T ]βα =  . ..  .. . am1 am2 11 m × n matrix of T with respect to α and β is  · · · a1n · · · a2n   ..  . .. . .  ··· amn Coordinate vectors. Let V be a finite-dimensional vector space, and let α = {v1 , . . . , vn } be a basis for V . For any v ∈ V , the coordinate vector of v with respect to α is the n-entry column vector   a1  ..  [v]α =  .  ∈ Rn , where a1 , . . . , an ∈ R are the unique scalars such that v = a1 v1 + · · · + an vn . an (The term “coordinate vector” and the notation [v]α and can vary from textbook to textbook. If a comps problem needs to refer to coordinate vectors, the problem itself will define the notation and terminology.) • Key Fact 1: For any v ∈ V , the coordinate vectors [v]α ∈ Rn and [T (v)]β ∈ Rm are related via the matrix [T ]βα ∈ Mm×n (R) by the equation [T ]βα [v]α = [T (v)]β . • Key Fact 2: If X is another vector space, with basis γ = {x1 , . . . , xp }, and if U : W → X is linear, then the m × n matrix [T ]βα , the n × p matrix [U ]γβ , and the m × p matrix [U T ]γα are related by matrix multiplication: [U T ]γα = [U ]γβ [T ]βα . 13 (January 2009) Let M2×2 be the vector space of 2 × 2 matrices with real entries. It is a fact that dim M2×2 = 4, and that {e1 , e2 , e3 , e4 } is a basis for M2×2 , where         1 0 0 1 0 0 0 0 e1 = , e2 = , e3 = , e4 = . 0 0 0 0 1 0 0 1   −1 7 Consider the function T : M2×2 → M2×2 given by T (A) = A . 3 4 (a) Prove that T is linear. (b) Find the matrix for T with respect to the basis {e1 , e2 , e3 , e4 }.  −1 Solution. For ease of notation, define C = 3 (a): Given A, B ∈ M2×2 and c ∈ R, we have  7 and α = {e1 , e2 , e3 , e4 }. 4 T (A + cB) = (A + cB)C = AC + (cB)C = AC + c(BC) = T (A) + cT (B) (b): We compute      1 0 −1 7 −1 7 T (e1 ) = = = −1e1 + 7e2 + 0e3 + 0e4 , 0 0 3 4 0 0      0 1 −1 7 3 4 T (e2 ) = = = 3e1 + 4e2 + 0e3 + 0e4 , 0 0 3 4 0 0      0 0 −1 7 0 0 T (e3 ) = = = 0e1 + 0e2 − 1e3 + 7e4 , 1 0 3 4 −1 7      0 0 −1 7 0 0 T (e4 ) = = = 0e1 + 0e2 + 3e3 + 4e4 0 1 3 4 3 4   −1 3 0 0  7 4 0 0  . Putting these coefficients down columns, the matrix is [T ]α α = 0 0 −1 3 0 0 7 4 Comment. Make sure you understand why the matrix for T in part (b) is 4 × 4, not 2 × 2. 12 QED (a) Here are two problems you should try yourself. See also Example 17 later. 14 (March 2013) Let P2 be the vector space consisting of all polynomials of degree at most 2, and let B = {1, x + 1, x2 + x + 1}, which is a basis for P2 . Let T : P2 → P2 be the linear transformation defined by the equation T (f ) = f − f 0 . Find the matrix of T with respect to B.   1 Answer: 0 0 −1 1 0 1 −2 1 15 (January 2014) Suppose T : V → V is a linear transformation, B = {b1 , b2 , b3 } is a basis for V , and the matrix representation of T with respect to B is   2 3 5 A =  7 11 −3 . −1 19 0 Determine T (2b1 + 4b3 ) as a linear combination of b1 , b2 , and b3 . Answer: 24b1 + 2b2 − 2b3 Invertible maps. A linear transformation T : V → W is invertible if it has an inverse, i.e., if there is another linear map T −1 : W → V such that T (T −1 (w)) = w for all w ∈ W , and T −1 (T (v)) = v for all v ∈ V . Know that the following are equivalent: • T is invertible (i.e., T has an inverse T −1 : W → V ). • T is one-to-one and onto. In addition, recall that one of the earlier two-out-of-three theorems (see page 8 of this Guide) says the following: if either dim(V ) < ∞ or dim(W ) < ∞ (or both), then T being invertible is equivalent to each of the following statements: • dim(V ) = dim(W ) and T is one-to-one. • dim(V ) = dim(W ) and T is onto. Facts: Suppose T : V → W is invertible. Then: • Its inverse T −1 is unique. • Its inverse T −1 is also invertible, and (T −1 )−1 = T . • If U : W → X is also invertible, then U T : V → X is invertible, and (U T )−1 = T −1 U −1 . Invertible matrices. A square matrix A ∈ Mn×n (R) is invertible if it has an inverse, i.e., if there is another matrix B ∈ Mn×n (R) such that AB = I and BA = I. Know that the following are equivalent: • A is invertible • The columns of A are linearly independent. • The rows of A are linearly independent. • The columns of A together span Rn . • The rows of A together span Rn . • The columnspace (i.e., range or image) of A is all of Rn . • The nullspace (i.e., kernel) of A is {0}. • rank(A) = n. • nullity(A) = 0. • det(A) 6= 0. • λ = 0 is not an eigenvalue of A. Warning: the equivalencies above are only for square matrices. Nonsquare matrices are never invertible. 13 Facts: Suppose A ∈ Mn×n (R) is invertible. Then: • Its inverse A−1 is unique. • Its inverse A−1 is also invertible, and (A−1 )−1 = A. • If B ∈ Mn×n (R) is invertible, then AB is invertible, and (AB)−1 = B −1 A−1 . Know how to compute inverses:     1 d −b a b . • A= is invertible if and only if ad − bc 6= 0, in which case A−1 = c d ad − bc −c a • For 3 × 3 and larger matrices, use the Gauss-Jordan method (row reduction), as in the next example. (February 2006) Compute the inverse of the  2 A = 1 4 Check your answer by matrix multiplication. 16 matrix  1 1 0 0 . 1 2 Solution. We  ↓ 2 1 ↑  1 0 4 1 So A−1 do row reduction:     1 1 0 0 1 0 0 0 1 0 1 0 0 0  0 1 1 1 0 0 1 0  → −2R1  2 1 1 1 0 0  → 2 0 0 1 −4R1 4 1 2 0 0 1 −R2 0 1 2 0     1 0 0 0 1 0 0 1 0 1 0 0 1 −2 0  →  0 1 0 2 0 −1  → −R3  0 1 1 0 0 1 −1 −2 1 0 0 1 −1 −2 1   0 1 0 0 −1. Now we check by multiplication: = 2 −1 −2 1      2 1 1 0 1 0 1 0 0 0 −1 = 0 1 0 = I. AA−1 = 1 0 0  2 4 1 2 −1 −2 1 0 0 1  1 0 −2 0  −4 1 Comment: We only need to check either AA−1 = I or A−1 A = I. Because A is square, if one is true, then the other is automatically true. (Can you use the facts about invertible matrices to see why?) Change of basis. Let V be a finite-dimensional vector space, and let α = {v1 , . . . , vn } and β = {w1 , . . . , wn } both be bases for V . The change-of-basis matrix from α-coordinates to β-coordinates is the n × n matrix [I]βα , where I : V → V is the identity map I(v) = v. That is,   a11 a12 · · · a1n v1 = a11 w1 + a21 w2 + · · · + an1 wn  a21 a22 · · · a2n  v2 = a12 w1 + a22 w2 + · · · + an2 wn   [I]βα =  . , where  . . .. . .. .. ..   .. . an1 an2 ··· ann vn = a1n w1 + a2n w2 + · · · + ann wn As before, note that the grid of coefficients is flipped sideways to form the matrix. • For any v ∈ V , recall that [v]α ∈ Rn is the n-entry column vector of α-coordinates for v. We can compute the β-coordinate vector for v via the formula [v]β = [I]βα [v]α . β −1 • The change of coordinates matrix to change coordinates the other way is the inverse: [I]α . β = ([I]α ) 14 Change of basis matrices are often used when we can easily find out the matrix of some linear map T : V → W for one basis but we actually want the matrix of T for another basis, as in the following example. (January 2012) Let T : P2 → R2 , where P2 = {a + bt + ct2 : a, b, c ∈ R}, be defined by   p(1) T (p) = . p(2) You may assume that T is linear. (a) Find bases of the null space (kernel) and range of T .     1 −1 (b) Find the matrix representation of T with respect to the bases {1, t, t2 } and , . 0 1 17 Solution. Let α = {1, t, t2 }, which is the standard  basis for P2 , and let β = {e1 , e2 }, which is the 1 0 standard basis for R2 , where e1 = and e2 = . We have 0 1       1 1 1 2 T (1) = = 1e1 + 1e2 , T (t) = = 1e1 + 2e2 , T (t ) = = 1e1 + 4e2 , 1 2 4   1 1 1 and so the associated matrix is [T ]βα = . We now answer the questions: 1 2 4   1 1 1 (a): Row reduction gives echelon form , which has pivots in the first and second columns. 0 1 3     1 1 So the corresponding columns , of the original matrix form a basis for the range of T . 1 2 For the null space, name variables a, b, c for three columns; so c is free. The echelon form gives  the  2  b = −3c and so a = −b − c = 2c. That is, −3 is a basis for the null space of the matrix. But we   1   2 want the null space of T , and the coordinate vector −3 gives the polynomial 2 − 3t + t2 ∈ P2 (T ). 1 So {2 − 3t + t2 } is a basis for the null space of T .     1 −1 (b): Let γ = , . We want the matrix [T ]γα , and we already have [T ]βα . 0 1 Then the change-of-basis matrix from γ to standard (β) coordinates on R2 is   1 −1 [I]βγ = , 0 1 i.e., just the vectors of γ down the columns. [Make sure you understand why!] But we want the inverse [I]γβ , which by the 2 × 2 inverse formula is  
−1 1 1 [I]γβ = [I]βγ = . 0 1 So the desired matrix for T is      1 1 1 1 1 2 3 5 γ γ β [T ]α = [T ]α [I]β = = . 0 1 1 2 4 1 2 4 Comment. An alternate way to do part (b) would be to work out by hand that             1 −1 1 −1 1 −1 2 +1 , T (t) = 3 +2 , T (t ) = 5 +4 T (1) = 2 0 1 0 1 0 1 and then put these coefficients down columns to obtain the matrix above. 15 Here are two problems you should try yourself where change-of-basis matrices may be useful. Try each two ways: using change-of-basis matrices, and working things out as described in the Comment for Example 17 . (March 2012) Let P2 = {a + bt + ct2 : a, b, c ∈ R}, and suppose that T : R2 → P2 is a linear transformation which satisfies      1 3 T = 1 − 2t and T = t + 2t2 . 1 −1  8  (a) Find T . Answer: −1 + 5t + 6t2 . −4 (b) Is T one-to-one? Answer: Yes. (c) Is T onto? Answer: No. 18 19 (January 2015) Let V be the vector space of polynomials of degree at most 2, and let B = {1, x + 1, x2 + x + 1}, which is a basis for V . Suppose that T : V → V is a linear transformation, and the matrix of T relative to B is   1 2 3 3 0 −1 . 2 5 1 Find T (3x2 + x + 2). 4 Answer: 1 − 5x − 5x2 . Eigenvalues and Eigenvectors Let A be an n × n matrix, let λ ∈ R be a scalar, and let v ∈ Rn . We say that v is an eigenvector of A with eigenvalue λ if v 6= 0 and Av = λv. • To say “λ is an eigenvalue of A” means there exists v ∈ Rn r {0} such that Av = λv. • To say “v is an eigenvector of A” means that v 6= 0 and there exists λ ∈ R such that Av = λv. • An eigenvalue λ can be 0. • An eigenvector v cannot be 0. (By definition.) Side note: One can also define eigenvalues and eigenvectors for more general linear maps T : V → V , as well as to allow complex numbers, but neither of these topics appears on the comps exam. Computations. The characteristic polynomial of A is det(A − λI); that is, subtract the variable λ from each diagonal entry and take the determinant. The result is a polynomial of degree n in the variable λ. • Fact: The roots of the characteristic polynomial of A are precisely the eigenvalues of A. • The number of times that a given scalar λ shows up as a root of the characteristic polynomial is called the algebraic multiplicity of λ, or sometimes simply the multiplicity of λ. • The eigenspace Eλ of an eigenvalue λ is the nullspace N (A − λI) of the matrix A − λI. It consists of all the eigenvectors with eigenvalue λ, along with 0, which is not an eigenvector. • The eigenspace dimension dim(Eλ ) is sometimes called the geometric multiplicity of λ. • Fact: For any eigenvalue λ of A, we have 1 ≤ (geometric multiplicity of λ) ≤ (algebraic multiplicity of λ). 16 Thus, to find the eigenvalues and eigenvectors of a matrix A: 1. Compute the characteristic polynomial det(A − λI). 2. Find all roots of the characteristic polynomial; these are the eigenvalues. 3. For each eigenvalue λ, use row reduction to find a basis for the eigenspace Eλ = N (A − λI). Diagonalization. We say an n × n matrix A is diagonalizable if there is an invertible matrix P and a diagonal matrix D such that P −1 AP = D. The following are equivalent: • A is diagonalizable. • There is a basis for Rn consisting of eigenvectors of A. • All roots of the characteristic polynomial of A are real, and for each such root λ, (geometric multiplicity of λ) = (algebraic multiplicity of λ). In that case, the matrix P consists of the n linearly independent eigenvectors down the columns, and the diagonal matrix D has the eigenvalues along the diagonal, in the same order. Fact: If A has n distinct real eigenvalues, then A is certainly diagonalizable. (However, if A has repeated eigenvalues, it may or may not be diagonalizable.) 20 (January 2015) Let  5 A = 0 1 0 1 2  0 2. 4 (a) Find all eigenvalues of A. (b) Find, if possible, an invertible matrix P such that P −1 AP is diagonal, or show that no such matrix exists. Solution. (a): The characteristic polynomial of A is 5−λ 0 0 1−λ 2 0 1−λ 2 = (5−λ)(λ2 −5λ+4−4) = −λ(λ−5)2 , = (5−λ) det(A−λI) = 2 4−λ 1 2 4−λ so the eigenvalues are 0, 5, 5. (Or if you prefer, 0 and 5, where 5 has algebraic multiplicity 2.)   0 0 0 (b): Since λ = 5 has algebraic multiplicity 2, we focus on it first. We have A − 5I = 0 −4 2 . 1 2 −1  1 2 −1 Doing a quick row reduction (details omitted here) leads to the echelon form 0 2 −1, which has 0 0 0 2 pivots and hence nullity 3 − 2 = 1. That is, the nullspace of A − 5I has dimension only 1, which is strictly less than the algebraic multiplicity of λ = 5. Since there is an eigenvalue whose eigenspace has dimension strictly less than its algebraic multiplicity, the matrix A is not diagonalizable. That is, there does not exist an invertible matrix P such that P −1 AP is diagonal. Comment: When you’re trying to decide whether or not A is diagonalizable, hone in on any eigenvalues  of  algebraic multiplicity   at least 2. There’s no sense wasting your time finding the eigenvectors 0 0 −2 for λ = 0 and 1 for λ = 5 if it turns out — as it does here — that the matrix isn’t 1 2 diagonalizable. 17 (March 2010)  Find an 1 3 the matrix A = −3 −5 3 3 21 invertible matrix P and a diagonal matrix D such that P −1 AP = D for  3 −3. 1 Solution. First, we find the eigenvalues of A. The characteristic polynomial is det(A − λI) = 1−λ −3 3 3 3 −5 − λ −3 3 1−λ = (1 − t)(−5 − t)(1 − t) − 27 − 27 + 9(1 − t) + 9(1 − t) + 9(5 + t) = (1 − t)(−5 − t)(1 − t) + 9(−6 + 2 − 2t + 5 + t) = (1 − t)(−5 − t)(1 − t) + 9(1 − t) = (1 − t)[(t2 + 4t − 5) + 9] = (1 − t)(t2 + 4t + 4) = −(t − 1)(t + 2)2 . So the eigenvalues are 1, −2, −2. For the eigenvectors, since λ = −2 has algebraic multiplicity 2, we focus on it first. We have   3 3 3 1 1 1 A − (−2)I = −3 −3 −3. A quick row reduction (omitted here) gives echelon form 0 0 0, 3 3 3 0 0 0 which has one pivot and hence nullity 3 − 1 = 2, same as the algebraic multiplicity. From the echelon     1   1 form, we see that the eigenspace E−2 = N (A + 2I) has basis −1 ,  0    0 −1   0 3 3 For λ = 1, we have A − (1)I = −3 −6 −3. A quick row reduction (omitted here) gives echelon 3 3 0     1 1 0  1  form 0 1 1, and so the eigenspace E1 = N (A + I) has basis −1 .   1 0 0 0 Since each eigenvalue has eigenspace of the same dimension as its algebraic multiplicity, the matrix A is diagonalizable, with P −1 AP = D, where     1 1 1 −2 0 0 P = −1 0 −1 and D =  0 −2 0 0 −1 1 0 0 1 Comment 1 : This problem doesn’t explicitly ask you to find eigenvalues or eigenvectors. You have to recognize that finding P and D as requested means finding eigenvalues and eigenvectors. Comment 2 : We do not need to compute the inverse P −1 . Knowing that we have a basis of eigenvectors is enough to guarantee that P −1 AP = D. That said, make sure the order of eigenvalues along the diagonal of P corresponds to the order of eigenvectors in the columns of P . Warning: Don’t mix up the two adjectives “diagonalizable” and “invertible”. They have nothing to do with one another. For example:     1 1 1 1 is both diagonalizable and invertible. is diagonalizable but not invertible. 0 2 1 1  1 0  2 is invertible but not diagonalizable. 1  0 0  1 is neither invertible nor diagonalizable. 0 You should try confirming those facts about all four of the above 2 × 2 matrices. 18 Here are two diagonalization problems you should try yourself. 22 (March 2008) Let  2 A = 0 1 1 1 1  −2 0 . −1 (a) Find all eigenvalues of A. Answer: λ = 0, 1, 1. (b) Find an invertible matrix P such that P −1 AP is a diagonal matrix, or show thatthere is no 1 One Answer: P = 0 1 such matrix P . 23 −1 1 0 2 0. 1 (March 2009) Let M be the matrix  1 M = 2 1 −2 5 2 (a) Find the eigenvalues of M . (b) Is M diagonalizable? Why or why not? Note on  0 0 . 2 Answer: λ = 2, 3, 3. Partial Answer: No. : there is more than one possible correct answer for part (b). For example, we could  1 permute the three columns of P any way we choose. We could also replace the column 0 by any nonzero 1     2   −1 scalar multiple, and we could replace the other two columns  1  , 0 by any other basis for the (2  1 0 dimensional) eigenspace of λ = 1. For any of these changes, the columns would still be a linearly independent set of eigenvectors, making P −1 AP still a diagonal matrix. 22 Note on 23 , the “Answer” to part (b) doesn’t include the very important answer to “Why?” The full answer is very similar to part (b) of 20 above. In particular, a computation shows that the eigenvalue λ = 3 has algebraic multiplicity 2, but geometric multiplicity only 1. 19 5 Preparing for the Linear Algebra Exam Now that you have finished reading the content part of the Study Guide, what should you do next to prepare for the linear algebra exam? The key thing to keep in mind is that You need an active knowledge of linear algebra. Here are a some suggestions to help you achieve this. Read the Study Guide Actively. There are several places where the Study Guide asks you to do a problem yourself. Do so. For those problems, the Study Guide gives you the final answer so you can check your work. However, on the exam, we grade all of your work, not just the final answer. Read Your Notes and Your Linear Algebra Book. In many places in the Study Guide, we say “Know. . . ,” without stating the facts precisely. This is deliberate, since we want you to refer to your notes and your linear algebra book when studying for the exam. For example, this Study Guide doesn’t explain the strategy of row reduction or how to compute determinants (although some examples appear in the worked problems). You are expected to review how to do that yourself. Not everything covered in your linear algebra course is part of the exam. For example, orthogonality and the Gram-Schmidt process are important topics from linear algebra, but they are not covered on the comps exam. This Study Guide and the Syllabus for the Comprehensive Examination in Linear Algebra (Math 271/272 ) list the topics that you need to know. Know Basic Results and Definitions. Keep in mind that knowing the precise statements of definitions and basic theorems is essential. The adjective “precise” is important here. For example, if a problem asks you to define an eigenvector of a matrix A, writing just Av = λv will not get full credit. You need to state the whole definition: v ∈ Rn is an eigenvector of A if v 6= 0 and there exists λ ∈ R such that Av = λv. Study Old Exams and Solutions. The Department website has a collection of old Comprehensive Exams, many with solutions. It is very important to do practice problems. This is one of the key ways to acquire an active knowledge of linear algebra. However, there are two dangers to be aware of when using old exams and solutions: • Thinking that the exams tell you what to study. Every topic on the Syllabus and in this Study Guide is fair game for an exam question. • Reading the solutions. This is passive. To get an active knowledge of the material, do problems from the old exams yourself, and then check the solutions. The more you can do this, the better. Work Together, Ask Questions, and Get Help. Studying with your fellow math majors can help. You can learn a lot from each other. Faculty are delighted to help. Don’t hesitate to ask us questions and show us your solutions so we can give you feedback. The QCenter has excellent people who have helped many students in the past prepare for the Comprehensive Exam. Start Now. Properly preparing for the linear algebra exam will take longer than you think. Start now. 20